Question

If 0.750 L of argon at 1.50 atm and 177°C and 0.235 L of sulfur dioxide at 95.0 kPa and 63.0°C are added to a 1.00-L flask and the temperature is adjusted to 25.0°C, what is the resulting pressure in the flask?

Answer 1

Answer:

The resulting pressure in the flask is 0.93 atm

Explanation:

- Apply the Ideal Gas law in both cases to get the mols, of Ar and SO2.

- Once you know the mols, sum both to get the total mols in the mixture.

- Apply the Ideal Gas lawin the flask with the total mols to know the resulting pressure.

First: 0.750 L of argon at 1.50 atm and 177°C

T° C + 273 = T° K → 177°C + 273 = 450K

P .V = n . R . T

1.50 atm . 0.750 L = n . 0.082 L.atm/mol.K  . 450K

(1.50 atm . 0.750 L) /  (0.082 mol.K/L.atm  . 450K) = n

0.030 mols Ar = n

Be careful with the R units, the ideal gases constant

Let's convert kPa to atm.

101.33 kPa _____ 1 atm

95 kPa ________ (95 / 101.33) = 0.94 atm

T° C + 273 = T° K → 63°C + 273 = 336 K

0.94 atm . 0.235 L = n . 0.082 L.atm/mol.K . 336K

(0.94 atm . 0.235 L) / (0.082 mol.K/L.atm . 336K) = n

8.01X10⁻³ mols = n

0.030 mols Ar  + 8.01X10⁻³ mols SO₂ = 0.038 total mols in the mixture

1L . P = 0.038 mol . 0.082 L.atm / mol.K . 298 K

P = (0.038 mol . 0.082 L.atm / mol.K . 298 K ) / 1L

P = 0.93 atm